Problem 22 - General Chemistry: Principles and Modern Applications

Question

Predict in each case whether a reaction is likely to occur. If so, write a net ionic equation. (a) \(\mathrm{AgNO}_{3}(\mathrm{aq})+\mathrm{CuCl}_{2}(\mathrm{aq}) \longrightarrow\) (b) \(\mathrm{Na}_{2} \mathrm{S}(\mathrm{aq})+\mathrm{FeCl}_{2}(\mathrm{aq}) \longrightarrow\) (c) \(\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{aq})+\mathrm{AgNO}_{3}(\mathrm{aq}) \longrightarrow\)

Step-by-Step Solution

Answer

(a) No reaction occurs, hence no net ionic equation can be written. (b) Similarly, no reaction happens and no net ionic equation needed here. (c) The reaction here gives the net ionic equation as : \(2Ag^{+} (aq) + CO3^{2 -} (aq) \longrightarrow Ag2CO3 (s)\).

Step 1: Assess the reactivity

Looking at the reactivity series, silver (Ag) is less reactive than copper (Cu). Thus it is unlikely that copper would be displaced by silver in the first equation. Based on the reactivity series, there's no reaction.

Step 2: Write the net ionic equation for the reaction

The first equation doesn't proceed, so there's no reaction, hence no net ionic equation.

Step 3: Assess reactivity for second equation

The second reaction involves Na2S(aq) + FeCl2(aq). Considering the reactivity series sodium (Na) is more reactive than iron (Fe), implying sodium will not displace iron. So, no reaction occurs here.

Step 4: Write the net ionic equation for the second reaction

As the second reaction doesn't occur, no net ionic equation will be needed.

Step 5: Assess reactivity for third equation

The third equation is Na2CO3(aq) + AgNO3(aq). Based on the solubility rules, Ag+ ions would combine with CO3- ions to form the precipitate Ag2CO3.

Step 6: Write the net ionic equation for the third reaction

Here, a reaction occurs. Sodium and nitrate are spectator ions and do not participate in the reaction. This leaves: \(2Ag^{+} (aq) + CO3^{2 -} (aq) \longrightarrow Ag2CO3 (s)\). This is the net ionic equation for the third reaction.

Key Concepts

Reactivity Series

Understanding the reactivity series is essential when predicting chemical reactions, particularly displacement reactions. The reactivity series ranks metals by their ability to displace other metals from compounds, essentially listing them from most reactive to least reactive.
In the context of ionic equations, the reactivity series helps us predict if a metal will displace another from its compound. In the exercise given, we assess whether silver will displace copper and if sodium will displace iron. Here are some key points to consider:

• More Reactive Metals: Metals that are higher in the reactivity series will displace those that are lower.
• No Displacement: If the metal involved is less reactive than the one in the compound, no reaction will occur.

In our cases with silver and copper, and sodium and iron, no displacement occurs due to the respective positions of these metals in the series.

Solubility Rules

Solubility rules help determine whether a product of a reaction will precipitate or remain dissolved in water. They are a set of general guidelines that indicate the solubility of various ionic compounds in water.
An understanding of solubility rules is essential, especially when writing net ionic equations. In the third part of our exercise, the solubility rules guide us in predicting the formation of the insoluble compound, silver carbonate, Ag\(_2\)CO\(_3\). Here's how:

• Carbonates: Most carbonates, like silver carbonate, are insoluble except those involving Group 1 elements or ammonium.
• Spectator Ions: Ions that remain dissolved in solution and do not form part of the precipitate are known as spectator ions.

By applying these rules, we see that Ag\(_2\)CO\(_3\) precipitates from the reaction, with sodium and nitrate staying as spectator ions.

Precipitation Reactions

A precipitation reaction results in the formation of an insoluble product, or a precipitate, when two aqueous solutions are mixed. Understanding precipitation reactions is key to predicting what will occur in chemical equations.
In precipitation reactions, two dissolved substances react to form one or more solid products. In the provided exercise, we identify that because silver carbonate forms, a precipitation reaction occurs in the third equation. Here's why this matters:

• Driving Force: Precipitation is a driving force for reactions, as the formation of a solid is often energetically favorable.
• Identifying Precipitates: The use of solubility rules, as mentioned previously, helps us identify which compounds will precipitate.

This reaction involves Ag\(_2\)CO\(_3\) as the precipitate, emphasizing the interaction between Ag extsuperscript{+} and CO extsubscript{3} extsuperscript{2−} ions that result in the solid formation.